Clebsch-Gordan coefficients for all the normalized $S_\textrm{tot}^2$ eigenstates of a three spin system with spin $s_\textrm{tot} = 1/2$?

Question

The total spin operator is defined as $\vec{S}_\textrm{tot} = \vec{S}_1 + \vec{S}_2 + \vec{S}_3$ with $\vec{S}_1 = S\otimes\mathbb{I}\otimes \mathbb{I}$, $\vec{S}_2 = \mathbb{I}\otimes S\otimes \mathbb{I}$ and $\vec{S}_3 = \mathbb{I}\otimes\mathbb{I}\otimes S$. Furthermore, the square of the total spin is $S_\textrm{tot}^2 = \vec{S}_\textrm{tot}\cdot \vec{S}_\textrm{tot}$. My question is how to determine the normalized eigenstates of $S_\textrm{tot}^2$ with total spin quantum number $s_\textrm{tot} = 1/2$, that is, with eigenvalue $\frac{3\hbar^2}{4}$. I think I have to combine the two first spins, and then combine those with the third, but I'm not sure how!

Answer 1

Yes. Combine spins 1 and 2 to obtain $j_{12}=0$ and $j_{12}=1$. Then combine each with the remaining spin 3 to get the total spin -$1/2$ states.

Note that there are two sets of spin-$1/2$ states, "coming from" $j_{12}=0$ and $1$ respectively. The states are distinct (as you will see) even if the total spin are the same. Indeed one could, for "local" reasons of symmetry or whatever, use any orthogonal combinations of states in those sets (presumably having identical total projections $M_s$.)

Moreover, you get two mathematically equivalent but physically different sets if you couple first spins 2 and 3, i.e. if you start by building $j_{23}=0$ or $1$, and then couple spin 1. These latter two sets can be written as linear combos of the former two sets.